Review: step response of 1st order systems
Step response
in the s—domain
a
;
s(s + a)
in the time domain
¢
¡
1 − e−at u(t);
time constant
τ=
1
;
a
rise time (10%→90%)
Tr =
2.2
;
a
settling time (98%)
Ts =
4
.
a
c(t)
Initial slope =
1
=a
time constant
1.0
0.9
0.8
0.7
63% of final value
at t = one time constant
0.6
0.5
0.4
0.3
0.2
0.1
0
1
a
2
a
Tr
3
a
4
a
5
a
Ts
Figure 4.3
2.004 Fall ’07
steady state
(final value)
Figure by MIT OpenCourseWare.
Lecture 07 – Wednesday, Sept. 19
t
Review: poles, zeros, and the forced/natural responses
jω
system pole –
natural response
input pole –
forced response
system zero –
derivative &
amplification
jω
σ
−5
2.004 Fall ’07
0
σ
−5
Lecture 07 – Wednesday, Sept. 19
−2
0
Goals for today
•
•
Second-order systems response
– types of 2nd-order systems
• overdamped
• underdamped
• undamped
• critically damped
– transient behavior of overdamped 2nd-order systems
– transient behavior of underdamped 2nd-order systems
– DC motor with non-negligible impedance
Next lecture (Friday):
– examples of modeling & transient calculations for
electro-mechanical 2nd order systems
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
DC motor system with non-negligible inductance
Recall combined equations of motion
)
LsI(s) + RI(s) + Kv Ω(s) = Vs (s)
⇒
JsΩ(s) + bΩ(s) = Km I(s)
⎧ ·
µ
¶
µ
¶¸
LJ
Lb
K
Km
K
⎪
m
v
⎨
s2 +
+J s+ b+
Ω(s) =
Vs (s)
R
R
R
R
⎪
⎩ (Js + b) Ω(s) = K I(s)
m
Including the DC motor’s inductance, we find
⎧
Km
1
Ω(s)
⎪
⎪
=
⎪
¶
µ
¶
µ
⎪
Vs (s)
LJ
⎪
K
R
bR
+
K
b
⎪
m
v
⎪
⎪
+
s+
s2 +
⎪
⎪
J
L
LJ
⎪
⎪
⎨
¶
µ
b
⎪
⎪
s+
⎪
⎪
1
I(s)
⎪
J
⎪
⎪
=
⎪
¶
µ
¶
µ
⎪
Vs (s)
R
⎪
K
R
bR
+
K
b
m
v
⎪
⎪
⎩
+
s+
s2 +
J
L
LJ
2.004 Fall ’07
Quadratic polynomial denominator
Second—order system
Lecture 07 – Wednesday, Sept. 19
Step response of 2nd order system – large R/L
Overdamped
response
↔
dissipation >
energy storage
2.004 Fall ’07
L = 0.1H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 ,
R = 6Ω, b = 4kg · m2 · Hz; vs (t) = 30u(t) V.
Lecture 07 – Wednesday, Sept. 19
Step response of 2nd order system – large R/L
Overdamped
response
2.004 Fall ’07
L = 0.1H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 ,
R = 6Ω, b = 4kg · m2 · Hz; vs (t) = 30u(t) V.
Lecture 07 – Wednesday, Sept. 19
Step response of 2nd order system – large R/L
Overdamped
response
2.004 Fall ’07
L = 0.1H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 ,
R = 6Ω, b = 4kg · m2 · Hz; vs (t) = 30u(t) V.
Lecture 07 – Wednesday, Sept. 19
Comparison of 1st order and 2nd order overdamped
1st order
(L≈0)
2.004 Fall ’07
2nd order
(L=0.1H)
Lecture 07 – Wednesday, Sept. 19
Step response of 2nd order system – small R/L
Underdamped
response
↔
dissipation <
energy storage
2.004 Fall ’07
L = 1.0H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 ,
R = 6Ω, b = 4kg · m2 · Hz; vs (t) = 30u(t) V.
overshoot
Lecture 07 – Wednesday, Sept. 19
Comparison of 1st order and 2nd order underdamped
1st order
(L≈0)
2.004 Fall ’07
NO overshoot
2nd order
(L=1.0H)
Lecture 07 – Wednesday, Sept. 19
overshoot
Overdamped DC motor: derivation of the step response
Using the numerical values L = 0.1H, Kv = 6V · sec, Km = 6N · m/A, J =
2kg · m2 , R = 6Ω, b = 4kg · m2 · Hz we find
rad
Km
= 30
;
LJ
sec · V
b
R
+
= 62rad/sec;
J
L
bR + Km Kv
2
= 300 (rad/sec) .
LJ
Therefore, the transfer function for the angular velocity is
30
Ω(s)
= 2
.
Vs (s)
s + 62s + 300
A 2nd —order system is overdamped
if the transfer function denominator
has two real roots.
We find that the denominator has two real roots,
s1 = −5.290Hz,
s2 = −56.71Hz
⇒
30
Ω(s)
=
.
Vs (s)
(s + 5.290)(s + 56.71)
To compute the step response we substitute the Laplace transform of the voltage
source Vs (s) = 30/s and carry out the partial fraction expansion:
Ω(s) =
3
3.3
0.3
900
= −
+
⇒
s(s + 5.290)(s + 56.71)
s s + 5.290 s + 56.71
¤
£
ω(t) = 3 − 3.3e−5.29t + 0.3e−56.71t u(t).
This is the function whose plot we analyzed in slides #5—8.
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Overdamped DC motor in the s-domain
jω
(2nd
system poles
order overdamped)
input
pole
σ
−56.71
−5.29 −5
0
1st order
system pole
jω
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Undamped DC motor: no dissipation
Consider the opposite extreme where the dissipation due to both the resistor and
bearings friction is negligible, i.e. R = 0 and b = 0. Using the same remaining
numerical values L = 0.1H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 , we
find
rad
Km
= 30
;
LJ
sec · V
b
R
+
= 0;
J
L
bR + Km Kv
2
= 180 (rad/sec) .
LJ
Therefore, the transfer function for the angular velocity is
30
Ω(s)
= 2
.
Vs (s)
s + 180
A 2nd —order system is undamped
if the transfer function denominator has a
conjugate pair of two imaginary roots.
The denominator has a conjugate pair of two imaginary roots,
s1,2 = ±j13.42Hz
⇒
30
Ω(s)
=
.
Vs (s)
(s + j13.42)(s − j13.42)
Again, the step response is found by partial fraction expansion:
Ω(s) =
900
5
5s
= −
⇒
s(s + j13.42)(s − j13.42)
s s2 + (13.42)2
ω(t) = [5 − 5 cos (13.42t)] u(t).
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Undamped DC motor in the s-domain
Natural frequency ωn = 13.42rad/sec.
Period T = 2π/ωn = 4.24sec.
(2nd
system poles
order undamped)
jω
j13.42
input
pole
σ
−5
1st order
system pole
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
0
−j13.42
Underdamped DC motor: small dissipation
Finally, let us return to what we previously labelled as “underdamped” case,
i.e. L = 1.0H, Kv = 6V · sec, Km = 6N · m/A, J = 2kg · m2 , R = 6Ω,
b = 4kg · m2 · Hz. The values of L, R are such that the dissipation in the system
is negligible compared to the energy storage capacity. We then find
rad
Km
=3
;
LJ
sec · V
b
R
+
= 8rad/sec;
J
L
bR + Km Kv
= 30 (rad/sec)2 .
LJ
Therefore, the transfer function for the angular velocity is
30
Ω(s)
= 2
.
Vs (s)
s + 8s + 30
A 2nd —order system is underdamped
if the transfer function denominator has a
conjugate pair of two complex roots.
This denominator has a conjugate pair of two complex roots,
s1,2 = −4 ± j3.74 rad/sec
⇒
30
Ω(s)
=
.
Vs (s)
(s + 4 + j3.74)(s + 4 − j3.74)
We will now develop the partial fraction expansion method for this case, aiming
to find the step response:
µ
¶
90
90
K 2 s + K3
K1
Ω(s) =
=
= 90
+ 2
.
s(s + 4 + j3.74)(s + 4 − j3.74)
s(s2 + 8s + 30)
s
s + 8s + 30
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Underdamped DC motor: small dissipation
Using the familiar partial fraction method, we can find K1 = 1/30, K2 = −1/30,
K3 = −8/30, therefore
¶
µ
s+8
1
−
.
Ω(s) = 3
s s2 + 8s + 30
To find the inverse Laplace transform, we rewrite the denominator as a complete
square plus a constant, and break down the numerator into the sum of the
same factor that appeared in the denominator’s complete square plus another
constant:
Ã
!
1
(s + 4) + 4
Ω(s) = 3
.
−
s (s + 4)2 + 14
If the complete square instead of (s+4)2 were of the form s2 , the inverse Laplace
transform would have followed easily from Nise Table 2.1:
"
#
¸
·
s+4
s+4
= L−1
L−1 2
2 = cos (3.74t) + 4 sin (3.74t) .
s + 14
s2 + (3.74)
To take the extra factor of 4 into account, we must use yet another property of
Laplace transforms, which we have not seen until now:
£
¤
L e−at f (t) = F (s + a).
(Nise Table 2.2, #4).
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Underdamped DC motor: small dissipation
We apply this “frequency shift” property as follows:
"
#
s+4
L−1
2 = cos (3.74t) + 4 sin (3.74t) ⇒
s2 + (3.74)
⇒ L−1
"
(s + 4) + 4
2
(s + 4) + (3.74)
2
#
= e−4t [cos (3.74t) + 4 sin (3.74t)] .
Combining all of the above results, we can finally compute the step response for
the angular velocity of the DC motor as
½
³
´¾
ω(t) = 3 − 3 e−4t [cos (3.74t) + 4 sin (3.74t)]
u(t).
With a little bit of trigonometry, which we leave to you to do as exercise, we
can rewrite the step response as
½
¾
ω(t) = 3 − 4.39e−4t cos (3.74t − 0.82) u(t).
So the step response of the 2nd —order underdamped system is characterized by
a phase—shifted sinusoid enveloped by an exponential decay.
This step response was analyzed in slides #9—10 of today’s notes.
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
What the real and imaginary parts of the poles do
c(t)
Exponential decay generated by real
part of complex pole pair
Sinusoidal oscillation generated by
imaginary part of complex pole pair
t
Figure by MIT OpenCourseWare.
ζ≡
Figure 4.8
2.004 Fall ’07
Damping ratio
Undamped (“natural”) period
1
.
2π Time constant of exponential decay
Note: the underdamped oscillation frequency is not the same as
the natural frequency!
Lecture 07 – Wednesday, Sept. 19
Underdamped DC motor in the s-domain
overshoot
(2nd
system poles
order undamped)
jω
−4 + j3.74
j3.74
σ
0
−4
input
pole
−4 − j3.74
2.004 Fall ’07
−j3.74
Lecture 07 – Wednesday, Sept. 19
The general 2nd order system
We can write the transfer function of the general 2nd —order system with unit
steady state response as follows:
ωn2
,
s2 + 2ζωn s + ωn2
where
• ωn is the system’s natural frequency, and
• ζ is the system’s damping ratio.
The natural frequency indicates the oscillation frequency of the undamped
(“natural”) system, i.e. the system with energy storage elements only and
without any dissipative elements. The damping ratio denotes the relative contribution to the system dynamics by energy storage elements and dissipative
elements. Recall,
ζ≡
Undamped (“natural”) period
1
.
2π Time constant of exponential decay
Depending on the damping ratio ζ, the system response is
• undamped if ζ = 0;
• underdamped if 0 < ζ < 1;
• critically damped if ζ = 1;
• overdamped if ζ > 1.
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
The general 2nd order system
c (t)
Undamped
2.0
1.8
1.6
1.4
Underdamped
1.2
Critically
damped
1.0
0.8
0.6
Overdamped
0.4
0.2
0
0.5
1
1.5
2
2.5
3
3.5
4
Figure by MIT OpenCourseWare.
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
t
The general 2nd order system
Images removed due to copyright restrictions.
Please see: Fig. 4.11 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
Nise Figure 4.11
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
The underdamped 2nd order system
ωn2
,
2
2
(s + 2ζωn s + ωn )
The step response’s Laplace transform is
ωn
K2 s + K3
K1
1
× 2
+
=
.
s s + 2ζωn s + ωn2
s
s2 + 2ζωn s + ωn2
We find
K1 =
1
,
ωn2
K2 = −
1
,
ωn2
K3 =
2ζ
ωn
Substituting and applying the same method of completing squares that we did
in the numerical example of the DC motor’s angular velocity response, we can
rewrite the laplace transform of the step response as
1
−
s
(s + ζωn ) + p
2
ζ
1−
(s + ζωn ) +
ζ2
ωn2
ωn
¡
p
1 − ζ2
1−ζ
2
¢
.
Using the frequency shifting property of Laplace transforms we finally obtain
the step response in the time domain as
"
#
³ p
´
³
´
p
ζ
sin ωn 1 − ζ 2 t .
1 − e−ζωn t cos ωn 1 − ζ 2 t + p
1 − ζ2
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
0<ζ<1
The underdamped 2nd order system
ωn2
,
2
2
(s + 2ζωn s + ωn )
0<ζ<1
Finally, using some additional trigonometry and the definitions
p
ωd = ωn 1 − ζ 2 ,
σd = ζωn ,
ζ
tan φ = p
1 − ζ2
we can rewrite the step response as
1
−
1
p
1 − ζ2
×
e−σd t
The definitions above can be re—written
ζ=
p
×
cos (ωd t − φ)
jω
ωn
σd
,
ωn
σ
− ζ ωn = − σd
X
Figure 4.10
2.004 Fall ’07
s-plane
θ
ωd
,
ωn
p
1 − ζ2
ωd
⇒ tan θ =
=
.
σd
ζ
1 − ζ2 =
2
+ jωn 1 − ζ = jωd
X
− jωn 1 − ζ2 = −jωd
Figure by MIT OpenCourseWare.
Lecture 07 – Wednesday, Sept. 19
The underdamped 2nd order system
ωn2
,
2
2
(s + 2ζωn s + ωn )
Finally, using some additional trigonometry and the definitions
σd = ζωn ,
p
ωd = ωn 1 − ζ 2 ,
ζ
tan φ = p
1 − ζ2
we can rewrite the step response as
1
−
1
p
1 − ζ2
×
e−σd t
×
cos (ωd t − φ)
c (t)
forced response,
sets steady state
Exponential decay generated by real
part of complex pole pair
Sinusoidal oscillation generated by
imaginary part of complex pole pair
Figure 4.10
Figure by MIT OpenCourseWare.
2.004 Fall ’07
t
Lecture 07 – Wednesday, Sept. 19
0<ζ<1
Transients in the underdamped 2nd order system
Peak time
Tp =
c(t)
π
p
.
ωn 1 − ζ 2
Percent overshoot (%OS)
!
Ã
ζπ
× 100
%OS = exp − p
1 − ζ2
cmax
1.02cfinal
c final
0.98cfinal
−ln (%OS/100)
⇔ζ= q
π 2 + ln2 (%OS/100)
0.9cfinal
0.1cfinal
Tr
Tp
Ts
Figure by MIT OpenCourseWare.
t
Settling time
(to within ±2% of steady state)
´
³
p
2
ln 0.02 1 − ζ
4
Ts = −
≈
.
ζωn
ζωn
(approximation valid for
0 < ζ < 0.9.)
Figure 4.14
2.004 Fall ’07
Lecture 07 – Wednesday, Sept. 19
Transient qualities from pole location in the s-plane
Images removed due to copyright restrictions.
Please see: Fig. 4.19 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
Recall
ζ=
p
ωd
,
ωn
p
1 − ζ2
ωd
⇒ tan θ =
=
.
σd
ζ
Nise Figure 4.19
2.004 Fall ’07
σd
,
ωn
Lecture 07 – Wednesday, Sept. 19
1 − ζ2 =